Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $q = \dfrac{r^2 - 8r + 16}{-3r - 9} \times \dfrac{r + 3}{7r - 28} $
Explanation: First factor the quadratic. $q = \dfrac{(r - 4)(r - 4)}{-3r - 9} \times \dfrac{r + 3}{7r - 28} $ Then factor out any other terms. $q = \dfrac{(r - 4)(r - 4)}{-3(r + 3)} \times \dfrac{r + 3}{7(r - 4)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ (r - 4)(r - 4) \times (r + 3) } { -3(r + 3) \times 7(r - 4) } $ $q = \dfrac{ (r - 4)(r - 4)(r + 3)}{ -21(r + 3)(r - 4)} $ Notice that $(r + 3)$ and $(r - 4)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ \cancel{(r - 4)}(r - 4)(r + 3)}{ -21(r + 3)\cancel{(r - 4)}} $ We are dividing by $r - 4$ , so $r - 4 \neq 0$ Therefore, $r \neq 4$ $q = \dfrac{ \cancel{(r - 4)}(r - 4)\cancel{(r + 3)}}{ -21\cancel{(r + 3)}\cancel{(r - 4)}} $ We are dividing by $r + 3$ , so $r + 3 \neq 0$ Therefore, $r \neq -3$ $q = \dfrac{r - 4}{-21} $ $q = \dfrac{-(r - 4)}{21} ; \space r \neq 4 ; \space r \neq -3 $